Discussion
If the stiffness of a system is quadrupled with mass unchanged, its natural period
Answer: D
Correct answer: halves
\[ T = 2\pi\sqrt{\frac{m}{k}} \quad\Rightarrow\quad T \propto \frac{1}{\sqrt{k}}, \] so quadrupling \( k \) divides the period by \( \sqrt{4} = 2 \). Stiffening a building therefore shortens its period, which usually moves it towards the plateau of the spectrum and increases the acceleration it attracts - a trade the designer must make deliberately.
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